User:GabrielaV

Introduction

Do you want to know how to contact me or find out some interesting things about me? [[1]]

Signals & Systems

Example

Find the first two orthogonormal polynomials on the interval [-1,1]

1. What is orthogonormal? [2]

2. What is orthogonal? [3]

3. What is a polynomial? [4]

        ${\mathbf {a}}$
        ${\mathbf {b}}t+c$

4. Now we can find the values for the unknown variables.

$<a|a>=\int _{-1}^{1}aadt=1$

${\mathbf {a}}={\sqrt {\frac {1}{2}}}$

$<bt+c|a>=\int _{-1}^{1}a(bt+c)dt=0$

${\mathbf {c}}=0$

$<bt+c|bt+c>=\int _{-1}^{1}(bt+c)^{2}dt=1$

${\mathbf {b}}={\sqrt {(}}{\frac {3}{2}})$

5. Now that we know what the first two orthogonormal polynomials!

Fourier Transform

As previously discussed, Fourier series is an expansion of a periodic function therefore we can not use it to transform a non-periodic funciton from time to the frequency domain. Fortunately the Fourier transform allows for the transformation to be done on a non-periodic function.

In order to understand the relationship between a non-periodic function and it's counterpart we must go back to Fourier series. Remember the complex exponential signal? [5]

$x(t)=x(t+T)=\sum _{k=-\infty }^{\infty }\alpha _{k}e^{\frac {j2\pi kt}{T}}$

where

${\mathbf {\alpha }}_{k}={1/T}\int _{-{T \over 2}}^{T \over 2}x(u)e^{-j2\pi ku \over T}du$

If we let

${\mathbf {T\to \infty }}$

The summation becomes integration, the harmoinic frequency becomes a continuous frequency, and the incremental spacing becomes a differential separation.

$\sum _{k=-\infty }^{\infty }\to \int _{-\infty }^{\infty }$

${{k \over T}\to f}$

${1/T}\to df$

The result is

$\lim _{T\to \infty }=\int _{-\infty }^{\infty }{\left[\int _{-\infty }^{\infty }x(u)e^{-j2\pi fu}du\right]}e^{j2\pi ft}df$

The term in the brackets is the Fourier transfrom of x(t)

${\mathcal {F}}[x(t)]=X(f)$

Inverse Fourier transform

$x(t)={\mathcal {F}}^{-1}[X(f)]$

How a CD Player Works

The first step on how a CD player works is that it reads $\sum _{n=-\infty }^{\infty }\ x(nt)\delta (t-nT)$ from the CD.

The data then goes through the Digital to Analog Converter and it is convolved with $\ p(t)$.

[[6]]

The result is $\sum _{n=-\infty }^{\infty }\ x(nt)\ p(t-nT)$ [[7]]

As you can see in the Frequency domain the final result does not appear to look like the original signal. Therefore we pass $P(f)\cdot {\frac {1}{T}}\sum _{m=-\infty }^{\infty }X(f-{\frac {m}{T}})$ through a low pass filter to knock out the high frequencies and then it will be outputted through the speaker.

2x Oversampling

The benefit of using oversampling is that this allows for more samples to be taken so you have an accurate digital signal which means better sound.

We have $\sum _{k=-\infty }^{\infty }\ x(nT)\delta (t-nT)$ [[8]]

but we want $\sum _{l=-\infty }^{\infty }(\sum _{m=-M}^{M}\ x({\frac {(l-m)T}{2}})h({\frac {mT}{2}}))\delta (t-{\frac {lT}{2}})$[[9]]

In order to get that we need to convolve it with $\sum _{m=-M}^{M}h({\frac {mT}{2}})\delta (t-{\frac {mT}{2}})$

now we convolve $\sum _{l=-\infty }^{\infty }(\sum _{m=-M}^{M}\ x({\frac {(l-m)T}{2}})h({\frac {mT}{2}}))\delta (t-{\frac {lT}{2}})$ with $p(2t)$ to then get $\sum _{l=-\infty }^{\infty }y({\frac {lT}{2}})p(2(t-{\frac {lT}{2}})$ and in the frequency domain it is ${\frac {1}{2}}P({\frac {f}{2}})\sum _{l=-\infty }^{\infty }y({\frac {lT}{2}})e^{-js\pi flT/2}$ [[10]] Now you only have to pass it through a low pass filter and you will have the original signal.

FINITE-IMPULSE REPONSE(FIR) FILTER

A FIR is one of two digital filters that is commonly used. This is that data $\sum _{n=-\infty }^{\infty }\ x(nt)\delta (t-nT)$ it is convolved with

$h(t)=\sum _{m=-M}^{M}h(mT/2)\delta (t-mT/2)$

the result is Failed to parse (syntax error): {\displaystyle \sum_{n=-\infty}^\infty\[\left sum_{m=-M}^M x(nT) h(mT/2)\right]\delta(t-nT-mT/2)}